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This question also was a part of my today's maths olympiad paper:

If squares of the roots of $x^4 + bx^2 + cx + d = 0$ are $\alpha, \beta, \gamma, \delta$ then prove that: $64\alpha\beta\gamma\delta - [4\Sigma \alpha\beta - (\Sigma \alpha)^2]^2 = 0$

I found value of $\alpha\beta\gamma\delta$ = $d^2$ and $\Sigma \alpha$ = -2b. How to move on ?

Vishwesh
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  • What is $\Sigma$? – Zev Chonoles Sep 01 '13 at 13:32
  • That means summation. For example: $\Sigma \alpha = \alpha + \beta + \gamma + \delta$ – Vishwesh Sep 01 '13 at 13:36
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    First of all, you should use the command \sum for that, not \Sigma. But more importantly, what summations do the expressions $$\sum \alpha,\qquad \sum\alpha\beta$$ represent? $\alpha$ and $\beta$ are specific numbers, not variables. It'd be like writing $$\sum 5$$ What does that mean? – Zev Chonoles Sep 01 '13 at 13:38
  • I will keep in mind that (sum). For the rest of the part: By $\sum \alpha$, the question means $\alpha + \beta + \gamma + \delta$ and by $\sum \alpha\beta = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta $ – Vishwesh Sep 01 '13 at 13:42

2 Answers2

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Set $y=x^2$ and find an equation for $y$.

To start with $y^2+by+cx+d=0$ so that $c^2y=c^2x^2=(y^2+by+d)^2=y^4+2by^3+(2d+b^2)y^2+2dy+d^2$ Whence $$y^4+2by^3+(2d+b^2)y^2+(2d-c^2)y+d^2=0$$

The equation for $y$ has roots which are the squares of the roots of the original equation. You can use Vieta's relations with this.

Mark Bennet
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Complete solution of the problem:

Given that α,β,γ & δ are squares of the roots of $x^4+bx^2+cx+d=0$ hence $\sqrt{α},\sqrt{β},\sqrt{γ}$ & $\sqrt{δ}$ are the roots of the given equation. Thus, we have

sum of the roots $$=\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ}=\frac{-B}{A}=\frac{0}{1}=0$$ &

sum of the products of two roots $$= \sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ}=\frac{C}{A}=\frac{b}{1}=b$$ now, taking the square of the first expression, we get $$(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})^2=0 $$ $$α+β+γ+δ+2(\sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ})=0$$

$$∑α+2b=0 $$ $∑α=-2b\tag1$

Now, the product of the roots $$=\sqrt{α} \sqrt{β} \sqrt{γ} \sqrt{δ}=\frac{E}{A}=\frac{d}{1}=d$$
$$ \sqrt{αβγδ}=d $$ $$ αβγδ=d^2 \tag2$$

Now, satisfying the given equation $x^4+bx^2+cx+d=0$ by its roots $\sqrt{α},\sqrt{β},\sqrt{γ}$ & $\sqrt{δ}$ one by one, we get the following $$\begin{align} α^2+bα+c\sqrt{α}+d=0\\ β^2+bβ+c\sqrt{β}+d=0\\ γ^2+bγ+c\sqrt{γ}+d=0\\ δ^2+bδ+c\sqrt{δ}+d=0\\ \end{align}$$ Now by adding above four expressions column-wise, we get $$α^2+β^2+γ^2+δ^2+b(α+β+γ+δ)+c(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})+4d=0$$

$$α^2+β^2+γ^2+δ^2+b(-2b)+c(0)+4d=0$$ $$α^2+β^2+γ^2+δ^2=2b^2-4d \tag3$$

Now by squaring both the sides of eq(1),we have $$(α+β+γ+δ)^2=(-2b)^2$$ $$α^2+β^2+γ^2+δ^2+2(αβ+αγ+αδ+βγ+βδ+γδ)=4b^2$$ By substituting the value from eq(3), we get $$2b^2-4d+2∑αβ=4b^2 $$
$$2∑αβ= 2b^2+4d $$
$$ 4∑αβ= 4b^2+8d \tag4$$

Now, by substituting all the corresponding values from eq(1), (2) & (4) in the LHS of the expression to be proved, we get

$$LHS=64αβγδ-[4∑αβ-(∑α)^2 ]^2=64d^2-[4b^2+8d-(-2b)^2 ]^2$$ $$ =64d^2-[8d]^2=64d^2-64d^2=0=RHS $$

Harish Chandra Rajpoot
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